Cos2x + 5cos(x+π) = -3?

Question

Find the specific solutions in the interval [0,2π) in radians in terms of π.

Answer 1

The trigonometric equation is cos 2x + 5cos(x + π) = - 3.

cos 2x + 5cos(π + x) = - 3

cos 2x + 5(- cos x) = - 3

cos 2x - 5cos x = - 3

Double angle formula : cos(2x) = 2cos² x - 1.

2cos² x - 1 - 5cos x = - 3

2cos² x - 1 - 5cos x + 3 = 0

2cos² x  - 5cos x + 2 = 0

Let, cos x = t.

Then, the trigonometric equation is 2t² - 5t + 2 = 0.

By factor by grouping.

2t² - 4t - t + 2 = 0

2t(t - 2) - 1(t - 2) = 0

Factor : (t - 2)(2t - 1) = 0.

Apply zero product property.

t - 2 = 0  and  2t - 1 = 0

t = 2  and  t = 1/2

put, t = cos x.

cos x = 2  and  cos x = 1/2

• cos x = 2

cos x = cos(cos^{- 1}(2)).

⇒ x = cos^{- 1}(2).

It does not exists, since the range of the cosine function is [- 1, 1].

• cos (x) = 1/2.

cos (x) = cos(π/3)

The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.

⇒x = 2nπ ± (π/3)

If n = 0, x = 2(0)π + (π/3) and x = 2(0)π - (π/3) = π/3 and - π/3,

If n = 1, x = 2(1)π + (π/3) and x = 2(1)π - (π/3) = 2π + π/3 and 2π - π/3 = 7π/3 and 5π/3.

⇒ x = π/3 and x = 5π/3 in the interval [0, 2π).

Therefore, the solutions of the given equation are x = π/3 and x = 5π/3 in the interval [0, 2π).